# Using Diodes As Series Noise Clippers

Frequently, noise in signals can cause unwanted triggering of sensitive circuits. If the noise voltage (Vn) is smaller than a diode forward voltage drop (Vf) and the signal voltage (Vs) is larger, a pair of parallel connected, opposite polarity (cathode of one attached to the anode of the other) diodes in series with a suitable resistor (R) can be used to eliminate the unwanted part of the input signal. Since the noise voltage is not large enough to forward bias either diode between signal peaks the output will be zero, producing a dead band of +/- Vf around ground level. When Vs forward bias either diode the output voltage will be (Vs – Vf)

If Vn is too large for ordinary diodes, two Zener diodes placed cathode to cathode are often substituted. The Zener voltage (Vz) is selected to be greater than the noise voltage. When the input signal goes positive, one diode is forward-biased, while the other enters breakdown mode. When the signal is negative, the roles are reversed. The dead zone is +/- (Vf + Vz), only signals greater than this will be passed to the output.

Since the voltage drop across the two diodes is subtracted from the input signal, the resulting output is (Vs – Vf – Vz).

For example, assume we have a +/- 6V input signal with noise amplitude of +/- 2V:
1) Find an appropriate Zener diode and calculate the value of R.
2) Calculate the amplitude of the output signals.

Given that Vz> 2V, a review of manufacturer specification sheets informs a 1N746A with Vz = 3.3V.

The output signal (Vo) is therefore:
= +/- (Vs – Vf – Vz)
= +/- (6V – 0.7V – 3.3V)
= +/- 2V.

In the absence of a load across R, the series resistor must pass enough current to keep the diode conducting when a signal is present. From the data sheet, Zener test current (Izt) is given as 20mA at Vz.

1) To ensure that the current through the resistor (Ir) is greater than the Zener breakdown or knee current, set it to approximately 1/4 X Izt = 1/4 X 20mA = 5mA.
2) Make the voltage drop across the resistor (Vr) equal to +/- 2V.

Now, the resistor value (R) will be equal to Vr / Ir = 2 / 5mA = 400 ohms. The nearest standard value would be 390 ohms. Therefore, power dissipation in R will be equal to (Vo X Vo) / R = (2V X 2V) / 390 ohms = 10.3mW.